Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

example: 12.11-3
A parallel RLC circuit with R 
→
∞
has an initial voltage of V
o
V across the capacitor and I
o
A in the 
inductor at t 
=
0
-
. Find expressions for all variables under free-response conditions.
Solution
A series RLC circuit with R 
=
0 and a parallel RLC circuit with R 
→
∞
will be the same circuit and 
hence the free-response in this case is same as the free-response in the case of undamped series RLC 
circuit. It is given in Eqn. 12.2-3 and is reproduced in the following.
v t
V
LI
C
t
t
i t
i t
I
CV
o
o
n
C
L
o
o
( )
cos(
)
( )
( )
s
=
+
−
≥
= −
= −
+
+
2
2
2
2
0
w
f
A for 
L
iin(
)
tan
w
f
f
n
o
o
t
t
I LC
V
−
≥
=








+
−
A for
where 
0
1
The ratio of amplitude of voltage to amplitude of current is 
L
C
. Total energy storage in the circuit 
will be 
LI
CV
o
o
2
2
2
2
+
J and it will remain constant at that value.
example: 12.11-4
Find the unit step response of a current excited parallel RLC circuit with L 
=
1 mH, C 
=
1000 
m
F and 
R 
=
 2.5 
W
.
Solution
Undamped natural frequency 
rad/s, critical resist
w
n
LC
=
=
1
1000
aance
Damping factor,
Natural fr
=
=
∴
=
=
∴
1
2
0 5
0 5 2 5 0 2
L
C
.
. / .
.
Ω
x
eequencies (
and
= − ±
−
= −
+
−
−
x
x w
j
j
j
n
1
200
979 8
200
979 8
2
)
.
.

---

TheParallel
RLC
Circuit

12.49
Capacitor behaves as a short-circuit and inductor behaves as an open-circuit under DC steady-state 
condition. Therefore, the 1 A current in the unit step will go through the inductor under steady state. 
Hence, the solution for i
L
(t) can be assumed as
i t
e
A
t A
t
i
i
L
t
L
( )
( sin
.
cos
. )
( )
= +
+
=
−
+
1
979 8
979 8
0
0
200
1
2
with 
and 
L
L
o
t
V L
( )
(
)
’
0
0
+
=
=
/
The initial conditions are zero valued since it is a step-response problem. We get the two equations 
needed to solve for A
1
and A
2
by applying initial conditions. They are A
A
A
2
1
2
1
979 8
200
0
= −
−
=
and
.
Solving for A
1
and A
2
, A
1
=
-
0.204 and A
2
=
-
1
∴
= −
+
≥
= −
−
+
i t
e
t
t
t
L
t
( )
( .
sin
.
cos
. )
.
1
0 204
979 8
979 8
0
1 1
200
A for
00206
979 8
0 2
0
200
e
t
t
v t
L i t
e
t
L
−
+
−
−
≥
=
=
cos(
.
.
)
( )
( ( ))
’
rad A for
2200
0 204
979 8
0 2
979 8
0 2
1 0206
t
t
t
e
[ .
cos(
.
.
) sin(
.
.
)]
.
−
+
−
=
−
rad
rad
2200
200
979 8
0
2 5 0 255
979
t
R
t
t
t
i t
v t
e
sin
.
( )
( ) / .
.
sin
V for
≥
=
=
+
−
..
( )
( )
( )
.
cos(
.
8
0
1
1 0206
979 8
200
t
t
i t
i t
i t
e
t
C
L
t
A for
R
≥
= −
−
=
+
−
−−
−
=
−
+
−
−
0 2
0 255
979 8
0 051
979 8
9
200
200
.
)
.
sin
.
[ .
sin
.
cos
rad
e
t
e
t
t
t
779 8
979 8
0 05
0
200
. ]
cos(
.
.
)
t
e
t
t
t
=
+
≥
−
+
rad A for
1.4
1.2
0.8
0.6
0.4
0.2
–0.2
1
2
Capacitor current
Voltage across the circuit
Time(ms)
Resistor current
Inductor current
Volts
Amps
3
4
5
6
7
8
9
10 11 12 13 14 15
–0.4
–0.6
1
Fig. 12.11-4 
VoltageandcurrentwaveformsinExample:12.11-4
All the input current goes through the capacitor at t 
=
0
+
since neither the voltage across the circuit 
nor the current through inductor can become non-zero at that instant. Inductor current shows 52.7% 
overshoot which is the value predicted by Eqn. 12.7-1 for 
x
=
0.2.
12.11.2 
Frequency response of parallel 
RLC
 circuit 
The parallel RLC circuit and its phasor equivalent circuit are shown in Fig. 12.11-5.

---

12.50


SeriesandParallel
RLC
Circuits
–
(a)
v
(
t
)
i
R
(
t
)
i
L
(
t
)
i
C
(
t
)
C
sin 
t
R
L
+
ω
–
(b)
C
R
L
+
V
(
j
)
ω
I
L
(
j
)
ω
I
C
(
j
)
ω
I
R
(
j
)
ω
I
S
(
j
)
ω
Fig. 12.11-5 
Parallel
RLC
circuitanditsphasorequivalentcircuit
Almost the entire source current flows through the inductor at low frequency since inductor is 
a short-circuit at DC and low impedance for low frequency AC. Similarly, almost the entire source 
current passes through the capacitor at high frequencies since capacitor impedance approaches zero 
as frequency increases without limit. Thus, the magnitude response (i.e., gain) of inductor current 
must be a low-pass function. Magnitude response of capacitor current must be a high-pass function 
and that of resistor current (and hence that of circuit voltage v(t) ) must be a band-pass function. These 
frequency-response functions are obtained by applying current division principle to the parallel RLC 
circuit phasor equivalent circuit. 
I
j
I
j
j L
R
j C
j L
LC
j L R
LC
LC
j
RC
L
S
(
)
(
)
w
w
w
w
w
w
w
w
w
=
+
+
=
−
+
=
−
+
=
1
1
1
1
1
1
1
1
2
2
w
w
w
w
xw w
n
n
n
j
2
2
2
2
(
)
−
+
This ratio can be written in polar form as 
I
j
I
j
L
S
n
n
n
L
L
n
n
(
)
(
)
(
)
tan
w
w
w
w
w
x w w
f
f
xw w
w
=
−
+
∠
= −
−
2
2
2 2
2
2
2
1
4
2
where
22
2
−
w
rad
Thus, the frequency-response function for i
L
(t) in a parallel RLC circuit is found to be the same as 
the frequency-response function for v
C
(t) in series RLC circuit. It is a low-pass output.
The frequency-response of i
C
(t) and i
R
(t) are also obtained similarly.
I
j
I
j
j
j
R
S
n
n
n
n
n
n
(
)
(
)
(
)
(
)
w
w
xww
w
w
xw w
xww
w
w
x w w
=
−
+
=
−
+
2
2
2
4
2
2
2
2 2
2
2
2
∠
∠
= −
−
−
f
f
R
R
n
n
where
rad
p
xw w
w
w
2
2
1
2
2
tan

---

TheParallel
RLC
Circuit

12.51
I
j
I
j
j
j
C
S
n
n
n
n
L
(
)
(
) (
)
(
)
w
w
w
w
w
xw w
w
w
w
x w w
f
=
( )
−
+
=
−
+
∠
2
2
2
2
2
2 2
2
2
2
2
4
whhere
rad
f
p
xw w
w
w
L
n
n
= −
−
−
tan
1
2
2
2
The frequency-response of voltage developed across the circuit is the frequency-response function 
for i
R
(t) multiplied by R. It will be a band-pass function. Hence, a parallel RLC circuit with high Q 
factor (i.e., low 
x
factor) will work as a narrow band-pass filter if it is excited by a current signal 
and the voltage across the circuit is accepted as the output. And, that is the most frequently used 
application of a parallel RLC circuit. 
These frequency-response functions have been dealt with in detail in the context of series 
RLC circuit and nothing further need to be added. Whatever that has been stated with respect to 
capacitor voltage in series circuit can be applied directly to inductor current in the parallel circuit and
so on.
Resonance in Parallel RLC Circuit takes place when input frequency is 
w
n
. Under resonant 
condition the input admittance (and impedance) of Parallel RLC Circuit becomes purely resistive and 
equal to 1/R Siemens. This is so since that frequency susceptance of inductor and capacitor are exactly 
equal in magnitude and opposite in sign and they cancel each other when added. They do not cancel 
completely at any other frequency and hence the admittance of a parallel RLC circuit is a minimum of 
1/R at resonant frequency.
All the current from the source flows through R under resonance conditions. Thus, amplitude of 
voltage across the parallel combination is a maximum of R V (assuming unit amplitude for source 
current) at 
w
n
. The amplitude of current through capacitor at that frequency will then be 
w
n
RC V. The 
amplitude of current through inductor at resonant frequency will then be R/
w
n
L V. Thus the current 
amplification factor at resonance in a parallel RLC circuit, defined as the ratio of amplitude of current 
in capacitor or inductor to the amplitude of source current, is
=
=
=
=
=
w
x
n
L C
R
RC
RC
LC
Q
1
1
2
/
.
Thus, a high Q circuit will carry very high amplitude currents in L and C even when the source 
current amplitude is small if the source frequency is equal to or near about the circuit resonant 
frequency. These currents cancel themselves due to their phase opposition and they do not consume 
any portion of the source current. Entire source current flows through the resistance under resonant 
condition.